Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
> On 1/22/2022 5:42 PM, WM wrote:
> NOT conserving <--> NOT two-ended stepping

That is nonsense, and it will not become true by repeating it. Here is my proof that matheology will be kept alive in every case, even by acting dishonestly. (If Cantor's work is invalid, modern mathematics goes up in smoke. The investment is too great – if something's wrong we'll just change logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct 1996)]).

Assume that my matrix had a first column number zero containing all the indexes. Then the eighth term of Cantor's sequence would be this

OXXXXOOOOO...
OXXXOOOOOO...
OXOOOOOOOO...
OOOOOOOOOO...
OOOOOOOOOO...
OOOOOOOOOO...
OOOOOOOOOO...
OOOOOOOOOO...
XOOOOOOOOO...
....

Then you and every matheologian would not need your counterfactual stepping-order-argument but simply claim that all O's would be gathered in this first column and no O would remain in the other columns of the matrix and no O would disappear. If you are honest enough you will admit it, and if not I know it nevertheless.

But this escape is blocked by starting with the column containig the n/1.

Regards, WM

WM used his keyboard to write :
> Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
>> On 1/22/2022 5:42 PM, WM wrote:
>
>> NOT conserving <--> NOT two-ended stepping
>
> That is nonsense, and it will not become true by repeating it. Here is my
> proof that matheology will be kept alive in every case, even by acting
> dishonestly. (If Cantor's work is invalid, modern mathematics goes up in
> smoke. The investment is too great – if something's wrong we'll just change
> logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct
> 1996)]).
>
> Assume that my matrix had a first column number zero containing all the
> indexes.

Why?

On 1/23/2022 3:49 AM, WM wrote:
> Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
>> On 1/22/2022 5:42 PM, WM wrote:
>
>> NOT conserving <--> NOT two-ended stepping
>
> That is nonsense, and it will not become true by repeating it. Here is my proof that matheology will be kept alive in every case, even by acting dishonestly. (If Cantor's work is invalid, modern mathematics goes up in smoke. The investment is too great – if something's wrong we'll just change logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct 1996)]).
>
> Assume that my matrix had a first column number zero containing all the indexes. Then the eighth term of Cantor's sequence would be this
>
> OXXXXOOOOO...
> OXXXOOOOOO...
> OXOOOOOOOO...
> OOOOOOOOOO...
> OOOOOOOOOO...
> OOOOOOOOOO...
> OOOOOOOOOO...
> OOOOOOOOOO...
> XOOOOOOOOO...
> ...

wrong! you have an X in the bottom row! Fail. Also your first column is bad.

Cantor's Enumeration does it the right way.

<snip crap>

On 1/22/2022 5:46 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 22. Januar 2022 um 22:18:36 UTC+1:

>> P matches X and Y _element by element_ iff
>> for each x in X, there is one and only one ⟨x,y⟩ in P,
>> and
>> for each y in Y, there is one and only one ⟨x,y⟩ in P.
>
> Like for every transposition there is one X and one O
> at the beginnig and at the end.

{X,O} has a two-ended stepping-order.
A flock of sheep has a two-ended stepping-order.

Not all collections have a two-ended stepping-order.

>> Matching step-by-step has tighter requirements than
>> matching element-by-element (matching à la Cantor).
>
> One of both is enough to be sure:
> No X will ever double and no O will disappear.

In a collection with X and O, we can prove from the
existence of a two-ended stepping-order that
no X will ever double and no O will disappear.

However,
no two-ended stepping-order of ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1}
exists.

FromTheRafters schrieb am Sonntag, 23. Januar 2022 um 13:20:24 UTC+1:
> WM used his keyboard to write :
> > Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
> >> On 1/22/2022 5:42 PM, WM wrote:
> >
> >> NOT conserving <--> NOT two-ended stepping
> >
> > That is nonsense, and it will not become true by repeating it. Here is my
> > proof that matheology will be kept alive in every case, even by acting
> > dishonestly. (If Cantor's work is invalid, modern mathematics goes up in
> > smoke. The investment is too great – if something's wrong we'll just change
> > logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct
> > 1996)]).
> >
> > Assume that my matrix had a first column number zero containing all the
> > indexes.
> Why?

In oder to show the dishonesty of matheologians. In that case they could claim that all O's could be gathered in the zeroths column while all matrix elements could be indexed. Therefore they need not take the silly step to let the O's disappear. But in case of my example this is impossible. Sinced they cannot admit that Cantor's theory is blatant nonsense, they let the O's disappear.

Regards, WM

Jim Burns schrieb am Sonntag, 23. Januar 2022 um 19:50:25 UTC+1:
> On 1/22/2022 5:46 PM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 22. Januar 2022 um 22:18:36 UTC+1:
>
> >> P matches X and Y _element by element_ iff
> >> for each x in X, there is one and only one ⟨x,y⟩ in P,
> >> and
> >> for each y in Y, there is one and only one ⟨x,y⟩ in P.
> >
> > Like for every transposition there is one X and one O
> > at the beginnig and at the end.
> {X,O} has a two-ended stepping-order.
> A flock of sheep has a two-ended stepping-order.
>
> Not all collections have a two-ended stepping-order.

Whatever you try, the O's will not disappear, neither during no after the aprocess of indexing.

> > No X will ever double and no O will disappear.
> In a collection with X and O, we can prove from the
> existence of a two-ended stepping-order that
> no X will ever double and no O will disappear.

I can prove from the blocked drain that never an O will disappear.
>
> However,
> no two-ended stepping-order of ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1}
> exists.

Irrelevant. The only relevant fact is that no O will go down the drain or diappear in any other way.

Regards, WM

WM brought next idea :
> FromTheRafters schrieb am Sonntag, 23. Januar 2022 um 13:20:24 UTC+1:
>> WM used his keyboard to write :
>>> Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
>>>> On 1/22/2022 5:42 PM, WM wrote:
>>>> NOT conserving <--> NOT two-ended stepping
>>>
>>> That is nonsense, and it will not become true by repeating it. Here is my
>>> proof that matheology will be kept alive in every case, even by acting
>>> dishonestly. (If Cantor's work is invalid, modern mathematics goes up in
>>> smoke. The investment is too great – if something's wrong we'll just change
>>> logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct
>>> 1996)]).
>>>
>>> Assume that my matrix had a first column number zero containing all the
>>> indexes.
>> Why?
>
> In oder to show the dishonesty of matheologians.

Oh, I see. Isn't that a little self-deprecating?

On 1/23/2022 1:12 PM, WM wrote:
> Jim Burns schrieb am Sonntag, 23. Januar 2022 um 19:50:25 UTC+1:
>> On 1/22/2022 5:46 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 22. Januar 2022 um 22:18:36 UTC+1:
>>
>>>> P matches X and Y _element by element_ iff
>>>> for each x in X, there is one and only one ⟨x,y⟩ in P,
>>>> and
>>>> for each y in Y, there is one and only one ⟨x,y⟩ in P.
>>>
>>> Like for every transposition there is one X and one O
>>> at the beginnig and at the end.
>> {X,O} has a two-ended stepping-order.
>> A flock of sheep has a two-ended stepping-order.
>>
>> Not all collections have a two-ended stepping-order.
>
> Whatever you try, the O's will not disappear, neither during no after the aprocess of indexing.

red herrings

>
>>> No X will ever double and no O will disappear.
>> In a collection with X and O, we can prove from the
>> existence of a two-ended stepping-order that
>> no X will ever double and no O will disappear.
>
> I can prove from the blocked drain that never an O will disappear.

they are your leaky rubber sets, fix it yourself.

>>
>> However,
>> no two-ended stepping-order of ℕ⁺⨯ℕ⁺ and ℕ⁺⨯{1}
>> exists.
>
> Irrelevant. The only relevant fact is that no O will go down the drain or diappear in any other way.

distraction

>
> Regards, WM

fredag 21 januari 2022 kl. 15:17:39 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 10:25:13 UTC+1:
> > fredag 21 januari 2022 kl. 09:56:41 UTC+1 skrev WM:
>
> > > > Empty assertion, I can show for ALL numbers.
> > > Then show me the first O that leaves the board.
> > > >
> > I do not care for your "board"
> And the board does not care for you and whether you care. It is existing and shows that matheologians who claim that all hOles disappear are stupid.
> >
> > for any rational number n/m I can find a natural number to map to it.
> But mathematical conclusion finds no hOle hone, all hOles present before after. That proves you wrong.
>
> Regards, WM
Nope, I can find a natural number, you have shown nothing. All that is just to obfuscate and I am not buying it.

FromTheRafters schrieb am Sonntag, 23. Januar 2022 um 20:13:24 UTC+1:
> WM brought next idea :
> > FromTheRafters schrieb am Sonntag, 23. Januar 2022 um 13:20:24 UTC+1:
> >> WM used his keyboard to write :
> >>> Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
> >>>> On 1/22/2022 5:42 PM, WM wrote:
> >>>> NOT conserving <--> NOT two-ended stepping
> >>>
> >>> That is nonsense, and it will not become true by repeating it. Here is my
> >>> proof that matheology will be kept alive in every case, even by acting
> >>> dishonestly. (If Cantor's work is invalid, modern mathematics goes up in
> >>> smoke. The investment is too great – if something's wrong we'll just change
> >>> logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct
> >>> 1996)]).
> >>>
> >>> Assume that my matrix had a first column number zero containing all the
> >>> indexes.
> >> Why?
> >
> > In oder to show the dishonesty of matheologians.
> Oh, I see. Isn't that a little self-deprecating?

Irrelevant. There is only one relevant question here: Is there any mistake in

1,2,4,7,OOOOO...
3,5,8,OOOOOO...
6,OOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
9,OOOOOOOO...
....

reproducing the first eight terms of the sequence (k(m/n)) where

k = (m + n - 1)(m + n - 2)/2 + m ?

Regards, WM

sergio schrieb am Montag, 24. Januar 2022 um 01:31:05 UTC+1:
> On 1/23/2022 1:12 PM, WM wrote:

> > Irrelevant. The only relevant fact is that no O will go down the drain or disappear in any other way.
> distraction

The point is this: Is there any mistake in

1,2,4,7,OOOOO...
3,5,8,OOOOOO...
6,OOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
9,OOOOOOOO...
....

reproducing the first eight terms of the sequence (k(m/n)) where

k = (m + n - 1)(m + n - 2)/2 + m ?

Regards, WM

zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 07:17:02 UTC+1:
> fredag 21 januari 2022 kl. 15:17:39 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 10:25:13 UTC+1:
> > > fredag 21 januari 2022 kl. 09:56:41 UTC+1 skrev WM:
> >
> > > > > Empty assertion, I can show for ALL numbers.
> > > > Then show me the first O that leaves the board.
> > > > >
> > > I do not care for your "board"
> > And the board does not care for you and whether you care. It is existing and shows that matheologians who claim that all hOles disappear are stupid.
> > >
> > > for any rational number n/m I can find a natural number to map to it.
> > But mathematical conclusion finds no hOle gone, all hOles present before after. That proves you wrong.
> >
> Nope, I can find a natural number, you have shown nothing. All that is just to obfuscate and I am not buying it.

Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in

1,2,4,7,OOOOO...
3,5,8,OOOOOO...
6,OOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
9,OOOOOOOO...
....

reproducing the first eight terms of the sequence (k(m/n)) where

k = (m + n - 1)(m + n - 2)/2 + m ?

Regards, WM

On Monday, January 24, 2022 at 10:51:59 AM UTC+1, WM wrote:
>
> Is there any mistake in
>
> 1,2,4,7,OOOOO...
> 3,5,8,OOOOOO...
> 6,OOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> ...
>
> reproducing the first eight terms of the sequence (k(m/n)) where
>
> k = (m + n - 1)(m + n - 2)/2 + m ?

What a silly question.

First of all your "sequence" isn't a sequence. But I guess that you meant the sequence

((m(k), n(k))_(k e IN),

i.e. the sequence

((1, 1), (1, 2), (2, 1), (1, 3), ...).

Your "figure" contains the indexes of the first eight terms of this sequence at the "positions" given by the terms. But it also contains infinitely more elements.

måndag 24 januari 2022 kl. 10:52:13 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 07:17:02 UTC+1:
> > fredag 21 januari 2022 kl. 15:17:39 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 10:25:13 UTC+1:
> > > > fredag 21 januari 2022 kl. 09:56:41 UTC+1 skrev WM:
> > >
> > > > > > Empty assertion, I can show for ALL numbers.
> > > > > Then show me the first O that leaves the board.
> > > > > >
> > > > I do not care for your "board"
> > > And the board does not care for you and whether you care. It is existing and shows that matheologians who claim that all hOles disappear are stupid.
> > > >
> > > > for any rational number n/m I can find a natural number to map to it.
> > > But mathematical conclusion finds no hOle gone, all hOles present before after. That proves you wrong.
> > >
> > Nope, I can find a natural number, you have shown nothing. All that is just to obfuscate and I am not buying it.
> Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> 1,2,4,7,OOOOO...
> 3,5,8,OOOOOO...
> 6,OOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> 9,OOOOOOOO...
> ...
>
> reproducing the first eight terms of the sequence (k(m/n)) where
>
> k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

Care to actually bother not fucking obfuscating it?

We have a bijection, it is done. Get over it already

On 1/24/2022 3:51 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 23. Januar 2022 um 20:13:24 UTC+1:
>> WM brought next idea :
>>> FromTheRafters schrieb am Sonntag, 23. Januar 2022 um 13:20:24 UTC+1:
>>>> WM used his keyboard to write :
>>>>> Jim Burns schrieb am Sonntag, 23. Januar 2022 um 01:45:11 UTC+1:
>>>>>> On 1/22/2022 5:42 PM, WM wrote:
>>>>>> NOT conserving <--> NOT two-ended stepping
>>>>>
>>>>> That is nonsense, and it will not become true by repeating it. Here is my
>>>>> proof that matheology will be kept alive in every case, even by acting
>>>>> dishonestly. (If Cantor's work is invalid, modern mathematics goes up in
>>>>> smoke. The investment is too great – if something's wrong we'll just change
>>>>> logic. [D.K. Davis in "Cantor's transfinite numbers", sci.math (31 Oct
>>>>> 1996)]).
>>>>>
>>>>> Assume that my matrix had a first column number zero containing all the
>>>>> indexes.
>>>> Why?
>>>
>>> In oder to show the dishonesty of matheologians.
>> Oh, I see. Isn't that a little self-deprecating?
>
> Irrelevant. There is only one relevant question here: Is there any mistake in
>
> 1,2,4,7,OOOOO...
> 3,5,8,OOOOOO...
> 6,OOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> 9,OOOOOOOO...
> ...
>
> reproducing the first eight terms of the sequence (k(m/n)) where
>
> k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

yes mistakes, and you are doing it wrong. your 9 is in the wrong place, 9 goes here;

1,2,4,7,OOOOO...
3,5,8,OOOOOO...
6,9,OOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
.....

Fail. Try Again. This is simple stuff!

On 1/24/2022 3:52 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 07:17:02 UTC+1:
>> fredag 21 januari 2022 kl. 15:17:39 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Freitag, 21. Januar 2022 um 10:25:13 UTC+1:
>>>> fredag 21 januari 2022 kl. 09:56:41 UTC+1 skrev WM:
>>>
>>>>>> Empty assertion, I can show for ALL numbers.
>>>>> Then show me the first O that leaves the board.
>>>>>>
>>>> I do not care for your "board"
>>> And the board does not care for you and whether you care. It is existing and shows that matheologians who claim that all hOles disappear are stupid.
>>>>
>>>> for any rational number n/m I can find a natural number to map to it.
>>> But mathematical conclusion finds no hOle gone, all hOles present before after. That proves you wrong.
>>>
>> Nope, I can find a natural number, you have shown nothing. All that is just to obfuscate and I am not buying it.
>
> Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
>
> 1,2,4,7,OOOOO...
> 3,5,8,OOOOOO...
> 6,OOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> 9,OOOOOOOO...
> ...
>
> reproducing the first eight terms of the sequence (k(m/n)) where
>
> k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

yes mistakes, and you are doing it wrong. your 9 is in the wrong place, 9 goes here;

1,2,4,7,OOOOO...
3,5,8,OOOOOO...
6,9,OOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
.....

Fail. Try Again. This is simple stuff!

Fritz Feldhase schrieb am Montag, 24. Januar 2022 um 11:33:32 UTC+1:
> On Monday, January 24, 2022 at 10:51:59 AM UTC+1, WM wrote:
> >
> > Is there any mistake in
> >
> > 1,2,4,7,OOOOO...
> > 3,5,8,OOOOOO...
> > 6,OOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > ...
> >
> > reproducing the first eight terms of the sequence (k(m/n)) where
> >
> > k = (m + n - 1)(m + n - 2)/2 + m ?
>
> First of all your "sequence" isn't a sequence.

It is a sequence of configurations. Above you see the configurations accumulated till number 8. k is the natural number 1, 2, 3, ... the position of which is defined by the arguments (m, n).

> But I guess that you meant the sequence
>
> ((m(k), n(k))_(k e IN),

There are many ways to express that m and n are the co-ordinates of the natural number k. All together, k, m, n, are required to define the configuration till the kth.
>
> i.e. the sequence
>
> ((1, 1), (1, 2), (2, 1), (1, 3), ...).
>
> Your "figure" contains the indexes of the first eight terms of this sequence at the "positions" given by the terms.

From this figure you can read that the first eight indexes are distributed according to Cantor's description. This can be extended infinitely as Cantor has taught us.

> But it also contains infinitely more elements.

Not yet. It contains all sites defined by the co-ordinates m and n, but not yet their Belegung.

Regards, WM

zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 12:19:46 UTC+1:
> måndag 24 januari 2022 kl. 10:52:13 UTC+1 skrev WM:

> > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > 1,2,4,7,OOOOO...
> > 3,5,8,OOOOOO...
> > 6,OOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > 9,OOOOOOOO...
> > ...
> >
> > reproducing the first eight terms of the sequence (k(m/n)) where
> >
> > k = (m + n - 1)(m + n - 2)/2 + m ?
> >
> Care to actually bother not fucking obfuscating it?
>
> We have a bijection, it is done. Get over it already

That is not the question here. The question has been defined above.

Regards, WM

sergio schrieb am Montag, 24. Januar 2022 um 16:24:06 UTC+1:
> On 1/24/2022 3:51 AM, WM wrote:

> > There is only one relevant question here: Is there any mistake in
> >
> > 1,2,4,7,OOOOO...
> > 3,5,8,OOOOOO...
> > 6,OOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > OOOOOOOOO...
> > 9,OOOOOOOO...
> > ...
> >
> > reproducing the first eight terms of the sequence (k(m/n)) where
> >
> > k = (m + n - 1)(m + n - 2)/2 + m ?
> >
> yes mistakes, and you are doing it wrong. your 9 is in the wrong place, 9 goes here;

No, above you see the eighth term of the sequence of configurations. All indexes k from 9 on are attached to integer fractions n/1 with n = k.

Regards, WM

On 1/24/2022 11:03 AM, WM wrote:
> sergio schrieb am Montag, 24. Januar 2022 um 16:24:06 UTC+1:
>> On 1/24/2022 3:51 AM, WM wrote:
>
>>> There is only one relevant question here: Is there any mistake in
>>>
>>> 1,2,4,7,OOOOO...
>>> 3,5,8,OOOOOO...
>>> 6,OOOOOOOO...
>>> OOOOOOOOO...
>>> OOOOOOOOO...
>>> OOOOOOOOO...
>>> OOOOOOOOO...
>>> OOOOOOOOO...
>>> 9,OOOOOOOO...
>>> ...
>>>
>>> reproducing the first eight terms of the sequence (k(m/n)) where
>>>
>>> k = (m + n - 1)(m + n - 2)/2 + m ?
>>>
>> yes mistakes, and you are doing it wrong. your 9 is in the wrong place, 9 goes here;
>
> No, above you see the eighth term of the sequence of configurations. All indexes k from 9 on are attached to integer fractions n/1 with n = k.
>
> Regards, WM

there is your problem.

On Monday, 24 January 2022 at 13:03:53 UTC-4, WM wrote:
> sergio schrieb am Montag, 24. Januar 2022 um 16:24:06 UTC+1:
> > On 1/24/2022 3:51 AM, WM wrote:
>
> > > There is only one relevant question here: Is there any mistake in
> > >
> > > 1,2,4,7,OOOOO...
> > > 3,5,8,OOOOOO...
> > > 6,OOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > 9,OOOOOOOO...
> > > ...
> > >
> > > reproducing the first eight terms of the sequence (k(m/n)) where
> > >
> > > k = (m + n - 1)(m + n - 2)/2 + m ?
> > >
> > yes mistakes, and you are doing it wrong. your 9 is in the wrong place, 9 goes here;
> No, above you see the eighth term of the sequence of configurations. All indexes k from 9 on are attached to integer fractions n/1 with n = k.

Are you NUTS? You are trying to mimic Cantor's counting of the positive rational numbers, and you think the numbers in the top row, starting from 2, represent integers? How FUCKING STUPID can you actually be?

On 1/23/2022 2:12 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 23. Januar 2022 um 19:50:25 UTC+1:
>> On 1/22/2022 5:46 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 22. Januar 2022 um 22:18:36 UTC+1:

>>>> P matches X and Y _element by element_ iff
>>>> for each x in X, there is one and only one ⟨x,y⟩ in P,
>>>> and
>>>> for each y in Y, there is one and only one ⟨x,y⟩ in P.
>>>
>>> Like for every transposition there is one X and one O
>>> at the beginnig and at the end.
>>
>> {X,O} has a two-ended stepping-order.
>> A flock of sheep has a two-ended stepping-order.
>>
>> Not all collections have a two-ended stepping-order.
>
> Whatever you try, the O's will not disappear,
> neither during no after the aprocess of indexing.

Whatever you want to call it, "disappearing", or
"explosion", or "matheology", the X's and O's on n/d
can be covered by the X's from k/1 alone,
where k = (n+d-1)*(n+d-2)/2 + n

Let X or O be on n/d.

Is there an X on k/1 to cover it?

Yes.
k = (n+d-1)*(n+d-2)/2 + n

Is the X on k/1 where k = (n+d-1)*(n+d-2)/2 + n
required to cover more than one X or O?
On, let us say, n/d and n'/d' ?

No.

For integer n,d,n',d' >= 1 let
| | k = (n+d-1)*(n+d-2)/2 + n
| | k = (n'+d'-1)*(n'+d'-2)/2 + n'

| (n+d-1)*(n+d-2)/2 < (n+d-1)*(n+d-2)/2 + n
| | (n+d-1)*(n+d-2)/2 + n < (n+d-1)*(n+d-2)/2 + n+d

Let r = n+d

(r-1)*(r-2)/2 + s = ((r+1)-1)*((r+1)-2)/2 + 1

| (r-1)*(r-2)/2 < k
| | k < ((r+1)-1)*((r+1)-2)/2 + 1
| | k =< ((r+1)-1)*((r+1)-2)/2

Let s = n'+d'
Similarly,

| (s-1)*(s-2)/2 < k
| | k =< ((s+1)-1)*((s+1)-2)/2

Together,
| (r-1)*(r-2)/2 =< ((s+1)-1)*((s+1)-2)/2
| | (s-1)*(s-2)/2 =< ((r+1)-1)*((r+1)-2)/2

For r,s >= 2
(r-1)*(r-2)/2 < (s-1)*(s-2)/2 iff r < s

| r =< s+1
| | s =< r+1

Finishing,
| r = s
| n+d = n'+d'

Because k = k and
(n+d-1)*(n+d-2)/2 + n = (n'+d'-1)*(n'+d'-2)/2 + n'

| n = n'
| | d = d'

_Eppur si muove_

>>> No X will ever double and no O will disappear.
>>
>> In a collection with X and O, we can prove from the
>> existence of a two-ended stepping-order that
>> no X will ever double and no O will disappear.
>
> I can prove from the blocked drain that never
> an O will disappear.

Keep the drain blocked.
No X's or O's are mysteriously located in the dark.

The bathtub ℕ⁺⨯ℕ⁺ can be covered by part, ℕ⁺⨯{1}
because it is only two-ended-stepping-orderable
bathtubs which CANNOT be covered by part.

Your "proof" refers to two-ended-stepping-orderable
PARTS of the bathtub as though they are ALL the
bathtub. The parts are not all.

måndag 24 januari 2022 kl. 18:00:52 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 24. Januar 2022 um 12:19:46 UTC+1:
> > måndag 24 januari 2022 kl. 10:52:13 UTC+1 skrev WM:
>
> > > Try to approach this question with no regard to set theory: Is there any error in the distribution of indices in
> > > 1,2,4,7,OOOOO...
> > > 3,5,8,OOOOOO...
> > > 6,OOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > OOOOOOOOO...
> > > 9,OOOOOOOO...
> > > ...
> > >
> > > reproducing the first eight terms of the sequence (k(m/n)) where
> > >
> > > k = (m + n - 1)(m + n - 2)/2 + m ?
> > >
> > Care to actually bother not fucking obfuscating it?
> >
> > We have a bijection, it is done. Get over it already
> That is not the question here. The question has been defined above.
>
> Regards, WM

You are doing those things just to obfuscate, just accept that there is a bijection and get over it already

horand....@gmail.com schrieb am Montag, 24. Januar 2022 um 22:39:24 UTC+1:
> On Monday, 24 January 2022 at 13:03:53 UTC-4, WM wrote:

> > > > There is only one relevant question here: Is there any mistake in
> > > >
> > > > 1,2,4,7,OOOOO...
> > > > 3,5,8,OOOOOO...
> > > > 6,OOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > OOOOOOOOO...
> > > > 9,OOOOOOOO...
> > > > ...

> > All indexes k from 9 on are attached to integer fractions n/1 with n = k.
> You are trying to mimic Cantor's counting of the positive rational numbers, and you think the numbers in the top row, starting from 2, represent integers?

No, they represent the unit fractions 1/1, 1/2, 1/3, 1/4, ... and are indexed by the shown integers.

Regards, WM

Jim Burns schrieb am Dienstag, 25. Januar 2022 um 01:36:58 UTC+1:
> On 1/23/2022 2:12 PM, WM wrote:

> > Whatever you try, the O's will not disappear,
> > neither during no after the aprocess of indexing.
> Whatever you want to call it, "disappearing", or
> "explosion", or "matheology", the X's and O's on n/d
> can be covered by the X's from k/1 alone,
> where k = (n+d-1)*(n+d-2)/2 + n

No.
>
>
> Let X or O be on n/d.
>
> Is there an X on k/1 to cover it?
>
> Yes.
> k = (n+d-1)*(n+d-2)/2 + n

Nevertheless all O's remain. Therefore your argument shows that too less fractions can be indeXed.

> > I can prove from the blocked drain that never
> > an O will disappear.
> Keep the drain blocked.
> No X's or O's are mysteriously located in the dark.

When the indexing has been "completed", all O's remain and all are in the dark.
>
> Your "proof" refers to two-ended-stepping-orderable

My proof refers to the fact that no X will let pass an O --- and nothing else. It is like Leonidas at the Thermopyles.

Regards, WM